Question 3.4: A stress element has σx = 80 MPa and τxy = 50 MPa cw, as sho...

(a) In the semigraphical approach used here, we first make an approximate freehand sketch of Mohr’s circle and then use the geometry of the figure to obtain the desired information.

Draw the σ and τ axes first (Fig. 3–11b) and from the x face locate σ_{x} = 80 MPa along the σ axis. On the x face of the element, we see that the shear stress is 50 MPa in the cw direction. Thus, for the x face, this establishes point A (80, 50^{cw}) MPa. Corresponding to the y face, the stress is σ = 0 and τ = 50 MPa in the ccw direction. This locates point B (0, 50^{ccw}) MPa. The line AB forms the diameter of the required circle, which can now be drawn. The intersection of the circle with the σ axis defines σ_{1} and σ_{2} as shown. Now, noting the triangle ACD, indicate on the sketch the length of  he legs AD and CD as 50 and 40 MPa, respectively. The length of the hypotenuse AC is

τ_{1}=\sqrt {(50)^{2} + (40)^{2}}=64.0 MPa

and this should be labeled on the sketch too. Since intersection C is 40 MPa from the origin, the principal stresses are now found to be

σ_{1}= 40 + 64 = 104 MPa  and  σ_{2}= 40 − 64 = −24 MPa

The angle 2φ from the x axis cw to σ_{1} is

2φ_{p} = tan^{−1} \frac {50}{40} = 51.3◦

To draw the principal stress element (Fig. 3–11c), sketch the x and y axes parallel to the original axes. The angle φ_{p} on the stress element must be measured in the same direction as is the angle 2φ_{p} on the Mohr circle. Thus, from x measure 25.7° (half of 51.3°) clockwise to locate the σ_{1} axis. The σ_{2} axis is 90° from the σ_{1} axis and the stress element can now be completed and labeled as shown. Note that there are no shear stresses on this element.
The two maximum shear stresses occur at points E and F in Fig. 3–11b. The two normal stresses corresponding to these shear stresses are each 40 MPa, as indicated. Point E is 38.7° ccw from point A on Mohr’s circle. Therefore, in Fig. 3–11d, draw a stress element oriented 19.3° (half of 38.7°) ccw from x. The element should then be labeled with magnitudes and directions as shown.
In constructing these stress elements it is important to indicate the x and y directions of the original reference system. This completes the link between the original machine element and the orientation of its principal stresses.

(b) The transformation equations are programmable. From Eq. (3–10),

tan 2φ_{p} =\frac {2τ_{xy}}{σ_{x} − σ_{y}}                    (3–10)

φ_{p} =\frac{1}{2} tan^{−1} (\frac {2τ_{xy}}{σ_{x} − σ_{y}})=\frac {1}{2} tan^{−1} (\frac {2(−50)}{80})= −25.7◦, 64.3◦

From Eq. (3–8), for the first angle φ_{p} = −25.7◦,

σ =\frac {σ_{x} + σ_{y}}{2} +\frac {σ_{x} − σ_{y}}{2} cos 2φ + τ_{xy} sin 2φ               (3–8)

σ =\frac {80 + 0}{2} +\frac {80 − 0}{2}cos[2(−25.7)] + (−50) sin[2(−25.7)] = 104.03    MPa

The shear on this surface is obtained from Eq. (3–9) as

τ = −\frac {σ_{x} − σ_{y}}{2}sin 2φ + τ_{xy} cos 2φ   (3–9)

τ = −\frac {80 − 0}{2}sin[2(−25.7)] + (−50) cos[2(−25.7)] = 0   MPa

which confirms that 104.03 MPa is a principal stress. From Eq. (3–8), for φ_{p} = 64.3◦,

σ =\frac {80 + 0}{2} +\frac {80 − 0}{2}cos[2(64.3)] + (−50) sin[2(64.3)] = −24.03   MPa

Substituting φ_{p}= 64.3◦ into Eq. (3–9) again yields τ = 0, indicating that −24.03 MPa is also a principal stress. Once the principal stresses are calculated they can be ordered such that σ_{1} ≥ σ_{2}. Thus, σ_{1} = 104.03 MPa and σ_{2}= −24.03 MPa.

Since for σ_{1}= 104.03 MPa, φ_{p}= −25.7◦, and since φ is defined positive ccw in the transformation equations, we rotate clockwise 25.7° for the surface containing σ_{1}. We see in Fig. 3–11c that this totally agrees with the semigraphical method. To determine τ_{1} and τ_{1}, we first use Eq. (3–11) to calculate φ_{s} :

tan 2φ_{s} = −\frac {σ_{x} − σ_{y}}{2τ_{xy}}     (3–11)

φs =\frac {1}{2}tan^{−1} (−\frac {σ_{x} − σ_{y}}{2τ_{xy}})= \frac {1}{2}tan^{−1}(−\frac {80}{2(−50)})= 19.3◦, 109.3◦

For φ_{s}= 19.3◦, Eqs. (3–8) and (3–9) yield

σ =\frac {80 + 0}{2} +\frac {80 − 0}{2}cos[2(19.3)] + (−50) sin[2(19.3)] = 40.0 MPa

τ = −\frac {80 − 0}{2} sin[2(19.3)] + (−50) cos[2(19.3)] = −64.0 MPa

Remember that Eqs. (3–8) and (3–9) are coordinate transformation equations. Imagine that we are rotating the x, y axes 19.3° counterclockwise and y will now point up and to the left. So a negative shear stress on the rotated x face will point down and to the right as shown in Fig. 3–11d. Thus again, results agree with the semigraphical method. For φ_{s} = 109.3◦, Eqs. (3–8) and (3–9) give σ = 40.0 MPa and τ = +64.0 MPa.
Using the same logic for the coordinate transformation we find that results again agree with Fig. 3–11d.